MHB Mechanics- connected particles

[Shah 72]

Friction= 1/(sqroot12)×80 cos 30= 20N
I get two equations
T-20=8a and 120-T=12a
a=5m/s^2
I don't know how to calculate the time. Also with this value of acceleration tension value is wrong and the textbook ans is 84N

 

[Shah 72]

View attachment 11166
Friction= 1/(sqroot12)×80 cos 30= 20N
I get two equations
T-20=8a and 120-T=12a
a=5m/s^2
I don't know how to calculate the time. Also with this value of acceleration tension value is wrong and the textbook ans is 84N

I calculated the time
S=ut +1/2at^2
(2+1.5)= 1/2×5×t^2
I got t=1s
For q(b) iam getting the ans tension in the rope= 60N but the textbook ans is 84N. Pls help

 

[skeeter]

$Mg - T = Ma$
$T - mg(\sin{\theta} + \mu \cos{\theta}) = ma$
———————————————————————
$Mg - mg(\sin{\theta} + \mu \cos{\theta}) = a(M+m)$

$a = \dfrac{g[M-m(\sin{\theta}+ \mu \cos{\theta})]}{M+m} = \dfrac{10\left[12-8\left(\frac{3}{4}\right) \right]}{20}= 3 \, m/s^2$

$T = M(g-a) = 12(7) = 84$N

 

[Shah 72]

$Mg - T = Ma$
$T - mg(\sin{\theta} + \mu \cos{\theta}) = ma$
———————————————————————
$Mg - mg(\sin{\theta} + \mu \cos{\theta}) = a(M+m)$

$a = \dfrac{g[M-m(\sin{\theta}+ \mu \cos{\theta})]}{M+m} = \dfrac{10\left[12-8\left(\frac{3}{4}\right) \right]}{20}= 3 \, m/s^2$

$T = M(g-a) = 12(7) = 84$N

Thank you very much!